Find $\sin\left(255^\circ\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (Choice B) B $\dfrac{-\sqrt{3}-1}{2}$ (Choice C) C $\dfrac{-\sqrt{6}-\sqrt{2}}{4}$ (Choice D) D $-\dfrac{2}{3}$
The strategy First, we should rewrite the given angle $255^\circ$ as the sum or difference of two special angles. Then, we can use the sine addition or subtraction identities in order to evaluate $\sin\left(255^\circ\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $255^\circ$ We can rewrite $255^\circ$ as follows. $\begin{aligned}255^\circ&=120^\circ+135^\circ\end{aligned}$ In other words, $255^\circ$ is the sum of the special angles $120^\circ$ and $135^\circ$. Evaluating $\sin\left(255^\circ\right)$ Using the sine addition identity, we get the following. $\begin{aligned} \sin\left(255^\circ\right)&= \sin\left(120^\circ+135^\circ\right) \\\\\\ &= \sin \left(120^\circ\right) \cos \left(135^\circ\right) + \cos \left(120^\circ\right) \sin \left(135^\circ\right) \\\\\\ &=\left(\dfrac{\sqrt{3}}{2}\right) \left(-\dfrac{\sqrt{2}}{2}\right) + \left(-\dfrac{1}{2}\right) \left(\dfrac{\sqrt{2}}{2}\right) \\\\\\ &=\dfrac{-\sqrt{6}-\sqrt{2}}{4} \end{aligned}$ Summary $\sin\left(255^\circ\right) = \dfrac{-\sqrt{6}-\sqrt{2}}{4}$